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BULLETIN 
UNIVERSITY OF WASHINGTON 

ENGINEERING EXPERIMENT STATION 

E'^TGINEEIRING EXPERIMENT STATION SERIES BUEEETIN No. 13 



TENSIONS IN TRACK CABLES AND 

LOGGING SKYLINES 

The Catenary Loaded at One Point 

BY 



Samuel Herbert Anderson 
k 
Assistant Professor of Physics 




SEATTLE. WASHINGTON 

PUBLISHED QUARTERLY BY THE UNIVERSITY 

June, 1921 



Entered as second class matter, at Seattle, under the Act of July 16. 1894. 



CORRIGENDA. 






Page 7, equation {22) 



V(S + So)" ^- C"' — (S + So) 



(Vs-o + C- --So) 



(22) 



Page 8, equation (23) : — 



V So rC XX So X X 

y + V So-^c-| = ( e'~-^e^) -| ( e^— e^) 

2 2 



(23) 



Page 10, equation {38) : — 

3^0 .XX So X X 

y + yo = — (e~ +"e^) + — (e<="— e~) 
2 2 



(38) 



Page 15, equation {55) 



-y=— (yo— So)(e^— 1) 



(55) 



LIBRARY OF CONGRESS 

OOCHMCNTS DIVISION 

- -■ -■— ..-^.i -.-. , p- ■ , ,.., 1 



TENSIONS IN TRACK CABLES AND 
LOGGING SKYLINES 

The Catenary Loaded at One Point 



INTRODUCTION 

Two problems of the catenar>^, viz., the common catenary and 
the parabohc catenary, are generally treated in texts on Mechanics. 
A third problem, that of a catenary loaded at one point, is equally 
important, but either is not discussed at all or given a sketchy treat- 
ment that is far from satisfactory. This type of catenary is of con- 
siderable practical importance, as a conveying cableway or a cable 
tramway closely approximates in form to this curve. In the West- 
ern States such devices are much used in mining and logging opera- 
tions for transporting loads from higher to lower levels by gravity. 
In this paper the equations of the catenary loaded at one point will 
be derived and the properties of such a catenary discussed. It will 
also be shown how the tensions may be computed and what the 
condition is for maximum tension. 

It will be assumed that (1) the cable is perfectly flexible and 
(2) the load is fixed at one point, neither of which assumptions 
agree with practice. The cables in use are generally steel, sometimes 
two inches or more in diameter and very stiff in short lengths. 
However, when they are used in lengths of 2000 to 3000 feet and 
the vertical deflections are small compared to the length, the 
relative stiffness is negligible, and hence, the first assumption a 
legitimate one. As the load travels along, the conformation of the 
cable is constantly changing and likewise the tensions. But by 
assuming a fixed position of the load, it is possible to obtain the 
properties of the catenary for an instantaneous position of the load, 
from which the condition for maximum tension in the cable may be 
determined. To start with, it is assumed that the load is at any 
point, so that the treatment is general- 



Tensions in Track Cables and 




EIG. 1 



Logging Skylines 



Derivation of the Equation of the Catenary Loaded at One Point. 

Let OAB, figure I, represent a heavy, flexible cord or cable, 
fastened at A and B and supporting a known load W at O. Assume 
B to be at a higher horizontal level than A. At O there w^ill be three 
forces in equilibrium: (1) the weight, (2) the tension in the part 
of the cable OA which will be called T' and (3) the tension in OB 
which will be called T". Let </>'and </>" be the angles that T' and T" 
respectively make with the horizontal. By Lame's theorem we have 

W T' T" 



sin[180°— ((/>'+<^")] sin(90°+</>") sin(90°+</>') 
W T' T" 



= = (1) 

sm((f>'-\-ct>") cos^" cos<f>' 

Solving for W 

W=T'sin</.'+T'cos</>'tan<^" • (2) 

and 

W=T"tan(/,'cos</)"4-T"sin</>" (3) 

Resolving the forces vertically and horizontally, respectively, 
ve have 

W=T'sinc/.'+T"sinc^" (4) 

and 

T'cos</,'=T"cos</,"=Th (5) 

where Th is the horizontal component of the tension of the cable 
and is the same at all points for a given load and position of that 
load. Combining equations (2), (3), (4) and (5) we get 

2W=T'sin(/>' + T"sinc^"+Thtanc/>'+Thtan</." (6) 

W=Thtan(^'4-Thtanc/>" (7) 

It is evident that in form the cable is divided into two parts 
the point of division being at O, where the load is hung. From a 
mathematical consideration, this is a point of discontinuity which 
means that in passing from one portion of the cable to the other 
the change in the properties of the catenaiy with respect to the 
length is very abrupt at this point. It will be necessary, then, to 
consider the two parts separately and derive the equations for each 
part. 



6 Tensions in Track Cables and 

We will consider first the arc OB. Let P be any point of this 
arc, and let T be the tension at this point making an angle i// with 
the horizontal. The arc OP is in equilibrium under three forces : 
(1) the tension at O, which with respect to P is opposite and equal 
to T" of figure 1 ; (2) the weight of the cable between O and P 
acting at the center of gravity of OP; (3) the tension T at P. If 
s be the length of the cable from O to P measured along this arc and 
w the weight per unit length, then the second force is ws. Resolving 
these three forces vertically, we have 

Tsini/'=ws-[-T"sin</)" (8) 

and resolving horizontally 

Tcos./^Th=T"cos(/>" (9) 

Dividing equation (8) by (9) we have , 

dy ws 

tani/' = — = • \-tan4>" (10) 

dx T, 

This is the differential equation of the curve OB which is the con- 
formation the cable assumes under the action of the three forces. 
The solution of this gives the desired analytical equation. 

It is evident that in equation (10) w, Th, and tan</>" are all con- 
stants for a given cable, a given load and fixed position of that load. 
We will write for these three constants 

Tn 
c = - (11) 



and 



So=c tan</)" (12) 



Equation (11) conforms with the usual method of treatment of the 
common catenary, but it will be noticed that no assumption is made 
regarding the form of the curve OB. That is, c is considered noth- 
ing more than a constant at this stage of procedure. Introducing 
these two constants equation (10) becomes 

dy s+So 

-= (13) 

dx c 

Since s is a variable dependent upon x and y it is necessary to 
eliminate one of the three before the equation can be solved. This 
may be done by the relationship 



Logging Skylines 7 

ds2=dy^+dx2 (14) 

First eliminataing x we have 
dy s-|-So 



(15) 



ds V(s+So)-+c" 
the sohition of which is 



y=V(s+So)^+e'+A 

where A is a constant of integration. If we choose the origin at 
so that y^O when s^=0, then 



A= — Vs'o+c-' 
and (16) becomes 



y=V(s + So)= + c2' -Vs^' + e' . (17) 

On ehminating y between (13) and (14) we have 



ds V(s + So)^ + l' 



(18) 



dx 

the solution of which is 



log [(s-;rs)o+V(s-^s„)^'-fc--'|=^H-B (19) 

where B is a constant of integration. Using the same conditions as 
for evaluating the constant A, 



B= — log(so+ V+S"o+C=| ) 
Substituting this value of B (19) becomes 

(s + So)+V(s + So)^ + C^' 

_= log — — -^ (20) 

C So+Vs^o+C^' 

This may be written in the exponential form, 

^ (s + So)+V(s+So)- + C-| 

e~ = — -^ 

So+a/s'o+c^' 
By inverting equation (21) we get 
_^_ V (s+So)-+c-| — (s4-So) 
e=^ — ZZZZ^ 

(VS- + C^' —So) 

We may obtain an expression for y in terms of x and the con- 
stant c and So by adding (21) and (22)' and combining with (17). 



(21) 



(22) 



Tensions in Track Cables and 



V So |~C X X So 



y+Vso->e|= (e'-+eO=— (e>— eO (23) 

2 2 

By substracting- (21) from (22) we get a symmetrical expres- 
sion for s. 



V So ~t~C I X X So X X 

s+So = (e"^— e~) + — (e^+ e^) (24) 

2 2 

Equation (23) is the equation of the curve OB in rectangular 
co-ordinates, and (24) gives the length of the arc measured from O 
in terms of the abscissa and the constants So and c. These equations 
represent a general case of the catenary of which the common 
catenary is a special case. This may be seen from the following 
consideration. Suppose that tan<^" = 0, which could be attained by 
removal of the load W or by shifting to such a point that the arc 
OB became horizontal at O. Then by (12) So = and equations 
(23) and (24) reduce to 

^ X X 

y + c = — (e^+~e^) ' (25) 

2 
and 

c X _^ 
s = — (e^ — e^ (26) 

2 

which are the well known equations of the common catenary. 



In order to interpret the constants So and Vso^+c^l of equations 
(23) and (24) let us assume that the curve OB is extended to the 
left of O to a point C where the tangent is horizontal. This is rep- 
resented in figure 1 by the broken line curve CO. Let the weight 
per unit length of the ima"^inary cable be the same for the arc OB, 
and let the horizontal tension be Tu. Call the co-ordinates of 0> 
referred to C, p and q. The arc CO will be in equilibrium under 
three forces, Tn, the weight of the cable wr where r is the length of 
the arc CO, and the tension T" at O. Resolving these forces hori- 
zontally and vertically we have the following relations : 

Th=T"cos<^" (27) 

and 

T"sin<^"=wr (28) 

Dividing equation (28) by (27) we obtain 

dq r 
tanc/>" = — =— (29) 

dp c 



Logging Skyunes 



This is the differential equation of the arc CO, and is of the same 
form as that of the common catenary, so that the solution may be 
immeditely written. 



Vr=+c^'— c (30) 



p= clog^ V (31) 



) 

i c 

Since the curve COB is continuous the equations for any point 
referred to C as the origin will be of the same form as (30) and 
(31). Let x', y' be the co-ordinates of any point referred to C- Then 



y'=V(s+r)^ + c^' — c (32) 



f(s + r) + V(s + r)^ + c^'^ 

x' = c log^ ■ y (33) 

I c J 

where (s+r) is the length of the arc measured from C. 

Now let us transform the last two equations by the following 
relationship, 

y'=x+p 

y'=y+q 

and substituting values of p and q from (30) and (31). 

y=V(s+r)^+c=' - c + Vr' + c^^ + c (34) 



r(s + r) + V(s + r)^+c^' ^ 
I r — Vr^ + c^' ^ 



^=\og\ y (35) 



c L r — Vr^ + c^ 

Equations (34) and (35) must be identical with (17) and (20) 
respectively, since in both cases x, y are the co-ordinates of any 
point of the arc OB referred to O as an origin. Hence 

r = So 

That is, the constant So of equations (17) and (18) is the length 
of the fictitious arc CO. This means that the curve OB is a portion 
of a common catenary, the lowest point of zvhich is a distance So 
from O measured along the curve. While OB is the only part of 
this catenary that represents the conformation of a part of the 
cable, nevertheless, the part of the catenary CO has a physical in- 
terpretation which will be given later. 



10 Tensions in Track Cables and 

From a consideration of equations (17), (34), and (30) it is 
evident that 



Vso' + e' = Vr^ + c='=q + c (37) 

In the common catenary c is the distance from the lowest point to 
a horizontal line which may be called the directrix. Hence Vso^+c^' 
is the distance of O from this same directrix. We will call this yo. 
Making this substitution in equations (23) and (24) we get 

yo XX So X X 

y + vo = — (e"^ + e~) + — (e^— 'e~) (38) 

2 2 

yo X X So X X 

s + So = — ("e^ — e~) H (e~ +'e) (39) 

2 2 

Substituting the hyperbolic functions for exponential, these equa- 
tions become 

X X 

y + yo = yo cosh )- So sinh — (40) 

c c 

X X 

s + So = yosinh 1- So cosh — (41) 

c c 

These equations involve three constants, c, So, and yo, the last of 
which is dependent upon the other two. 



yo=Vso^+c-^' ' (42) 

All three are quantities of the fictitious arc CO ; c being the parame- 
ter, So the length of the arc, and yo the ordinate of the terminal 
point O measured from the directrix. By assigning different values 
to c, So, and yo (40) and (41) become the equations of a family of 
curves which represent the conformation the cable will take under 
various loads and horizontal tensions. 

In the same manner we may obtain equations of the arc OA 
which is the conformation assumed by the part of the cable to the 
left of the load W. The differential equation is, 

dy ws 

— = htanc/>' (43) 

dx T, 

which differs from (10) in that x and s are taken to left of O and 
fp' replaces cf>". Hence the analytical equations of the arc OA will be 



Logging Skylines 11 

the same as (40) and (41) except the constants So and yo will have 
different values. The constant. c is the same for the two curves, 
since it depends upon the horizontal tension and the weight of the 
cable per unit length, which are the same for all parts of. the cable. 
Hereafter when Soiand yoi are used in equations (40) and (41) 
they apply to the arc OA to the left of the origin and hence the 
values assigned to s and x are negative only ; and when Soo and yoo 
are used the above equations apply to the arc OB to the right of the 
origin and the values assigned to s and x are positive only. Since 
the constant c is the same for both arcs it is evident that OA and OB 
are arcs of the same common catenary having the parameter c. If 
in figure 1 the cui"ve DOA were shifted to the left and slightly up- 
ward until the point D coincided with C, AODCOB would form a 
continuous curve which is the common catenary from which OA and 
OB are taken to form the discontinuous curve AOB. The length 
of the arc DO is the constant Soi of the curve OA and the length of 
COO is the constant So2 of the curve OB. 



II. 

Determination of the Tensions. 

Arbitrarily we have set 

Th = wc (11) 

and found that c is a perameter of both curves OB and OA. 

The vertical components of the tensions T' and T" at O may be 
obtained by combining equations (5), (11) and (12). 

T'sin<^'=wSoi (44) 

T"sin<^"=wSo2 (45) 

Hence the vertical component of the tension in OA at O is equal to 
the weight of the fictitious arc OD ; and likewise, the tension in OB 
at is equal to the weight of OC. If these expressions for the 
vertical components of the tensions are substituted in equation (4), 
an expression is obtained giving the relation between the load W, 
the weight per unit length of the cable, and the fictitious arcs. 

W=w(so,+So,) ' (46) 

This shows that the constants Soi and Soo ai'e dependent upon the 
nature of the cable and the load. 



12 Tensions in Track Cabi^es and 



T'==wV e + So,^' =wyo, (47) 

T" = wVc^ + So/'=vvyo, . (48) 

The vertical component of the tension at any point in the curve 
OB is given by equation (8). Combining this with (45) we get 

T2sin,/^2=w(s+Soo) (49) 

And in hke manner we have for any point in the curve OA 

T'sin,/.,=w(s+SoO (50) 

Now (s+Soo) is the length of the arc COP measured from the 
lowest point of the fictitious arc OC, and hence the vertical compo- 
nent of the tension in the cable at any point is w times the length of 
the arc of the common catenary of zvhich OB or OA is a part. 

By combining (11) with (50) and (49) respectively we obtain 
the total tension at any point in OA or OB respectively. 

T,=w(y+yo,) (51) 

T,=w(y+yo,) (52) 

Hence the total tension at any point is w times the verticle distance 
of that point above the directrix of the common catenary. 

Equations (11), (44), (45), (47), (48), (49), (50), (51) 
and (52) give completely the tension in the cable. In every case 
the tension depends upon the weight per unit length, w, and one of 
the three constants, c, So, or yo. If these tensions are known, the 
constants can then be determined and by the aid of equations (40) 
and (41) the position of the cable determined. And conversely, 
if the constants can be determined from the data available and 
eouations (40) and (41), the tension for any point in the cable can 
be found. 

Equations (51) and (52) show that for a given load and given 
position of that load the tension in either part of the cable is great- 
est at the point where (y+yo) is a maximum, that is, at the point of 
fastening at the end of the cable. So that in future considerations 
of the effect of the load and of the position of the load the ten- 
sions at the ends only need be taken account of. At which of the 
two ends the tension is the greater depends upon the position of 
the load. If in equations (51) and (52) y^ is the verticle height of 
the lower end above the position of the load and y^ that of the 
upper end, y^ will be greater than Vj for any position of the load 



Page 12, insert before equation {47), 

The tensions T' and T" can be expressed in terms of w, c, So, 
and yo. Combining (11) with (44) and (45) respectively we have, 




FIG. 2 



14 Tensions in Track Cabi^es and 

along the cable, but yo2 will be considerably less than yo^ when the 
load is near the lower end and hence T, will be less that T^. But 
for the load near the middle of the span and any place between the 
middle and the higher end To will be greater than T^. (See fig- 
ure 2). 

In practice, information is wanted regarding the tensions in 
various parts of the cable for a certain load and different positions 
of that load, or more specifically, the maximum load the cable will 
carry. To answer the latter question equation (46) must be used 
which involves the constants s and Sq. It is desirable, then, to ex- 
press s in terms of other quantities on which it depends. This may 
be done by eliminating yo between equations (40) and (41). Writ- 
ing these in the following form 

X X 

y — Sosinh — = 3^0 (cosh — — 1) (40) 

c c 

X X 

s + So ( 1 — cosh—) = yosinh— (41 ) 

c c 

and dividing the first by the second we get. 



X 

y sinh c 




s 


So 

2 cosh X - 


-1 


2 


c 






which may be simplified to 




y 


s 




So = — coth(/) — 
2 


2 





(53) 
2 2 

where 

X 

^ = — (54) 

2c 

If Xj, yi are the coordinates of one end of the cable (the 
lower) referred to the position of the load as origin, s^ is the length 
of the portion of the cable extending from the load to the end, i.e., 
OA of figure 1. But the value of s^ depends upon the parameter c. 
An infinite number of catenaries may be drawn thru the points A 
and O, and the particular one which represents the conformation of 
the cable is determined by the value of c taken. In an unloaded 



Logging Skylines I5 

cable the value of c depends alone upon the horizontal tension 
which may be made anything desirable. In the loaded cable c may 
be computed from the value of the horizontal tension, but this lat- 
ter depends in part upon the load which is placed on the cable. A 
relationship between x, y, s and c, which is independent of the 
other constants Sq and yo, may be obtained which is of considerable 
use. Subtracting ecjuation (38) from (39) we get 

X 

s-y=-(yo-So)(e^-l) (55) 

and by adding 

X 

s+y=(yo+so)(e"-l) (56) 

On multiplying (55) by (56) and substituting 

c-=y^ — s% 
we have 

X X 

S2 y2=r=:c^(e'' 2-|-e'') 

^vhich may also be expressed in terms of the hyperbolic function, 

X 

Vs^ — y-|=2c sinh — (57) 

2c 

This may be thrown into a more useful form by substituting from 
equation (54). 

sinh (/> 
Vs^ — y^|=x (58) 

By this last equation the length of the cable between the load and 
terminal point may be quickly found for any value of c. To faciil- 
tate computation, it is desirable to have a curve in which sinh </> is 

1> 
plotted as ordinates against c^ as abscissas. Such a curve is given 
in figure 3. In the next section it will be shown how these computa- 
tions may be made. 

Another equation giving s in terms of x, y, and c may be obtain- 
ed by eliminating So between equations (40) and (41). 

x 

s = y + 2yotanh— (59) 

2c 



16 Tensions in Track Cables and 

III 

Practical Solution of the Problem. 

In sections I and II equations have been derived giving the 
relations among all the factors involved in stresses due to loading. 
But when an actual computation is undertaken, it is generally found 
that not sufficient data is available. Perhaps this may be best 
illustrated by taking a numerical problem, which was presented to 
the writer by a logging company. It is typical of those the engi- 
neers encounters. 

It is required to know the maximum load a cable will carry 
when used as a logging "sky-line" under the following conditions : 

1. Breaking strength — 112 tons. 

2. Weight of cable— 5.38 Ibs./linear ft. 

3. Horizontal span — 2500 ft. 

4. Vertical height of upper end above lower (tail spar tree 

above head spar tree) — 625 ft. 

5. Factor of safety — 3.5. 

The maximum tension allowable is 

112X2000 

= 64,000 lbs. 

3.5 

Calling this To and substituting in ecjuation (52) we find 

64,000 

y, + y^, = = 1 1 ,896 ft. 

5.38 

This may be used in equation (40) for finding So2, but before any 
computation is possible values of x, y, s and c are necessary. Since 
the value of Tg used above is the maximum that the cable is to 
stand, the value of x to be used should be the horizontal distance 
of the load from the upper end zvhere the tension is the greatest. 
In other words, before we can proceed further it is required to know 
the condition of maximum tension in the cable for a given load. 
So far no general method of treating this phase of the problem has 
been found, but it has been determined for a special case, which 
will be taken up in detail. 



Logging Skylines 



17 




FIG. 3 



18 Te;nsions in Track Cables and 

The first step necessary in this calculation is the assumption of 
some value of the parameter c. A common usuage* is to take the 
horizontal tension equal to one-fourth of the breaking strength of 
the cable, and the "deflection" of the cable when the maximum 
gross load is at the center to be one-twentieth of the span. By "de- 
flection" here is meant the vertical distance of the loaded point of 
the cable below the lower fixed end of the cable. 

Following thse two specifications we may find values of c, x 
and y. From equation (11) 

Th 112X2000 

= 10,409 



w 4 X 5.38 

One-twentieth of the span is 125 feet, so that 

y2=7S0 ft. 
X2=1250 ft. 

r.y equation (54) we find 

1250 



2X10409 



.06004 



sinh (f) 
From figure 3 =1.0012. 

Substituting these values of x,, ya, and sinh cj> in equation (58) 

we find 

S3=1458 ft. 

To find S] the same procedure is followed. 

Xi=1250 ft. 

yi=125 ft. 
sinh (/.= 1.0012 

The value of sinh <f> is the same for the two parts of the cable only 

for the middle of the span where x^^^Xo. This greatly simplifies 
calculations. It is found that 
Si=1257 ft. 

* Mark's Handbook of Mechanical Engineering, page 1160. 



Logging Skyunes 19 

The length of cable required for this span of 2500 feet is then 
5^+82=2715 ft. 

We may now compute Soi and So2, using equation (53). 

s„,=414 ft. 
s„2=5524 ft. 

The load which satisfies the above conditions if found by 
equation (46). 

W=5.38(414+5524)=31,946 lbs. 

To find the tensions at the two ends the constants yoi and joo 
must be found, using equation (42). 

yo,=10,419. 
yo2=l 1,784. 

Then by equations (51) and (52) the tensions are computed. 

T,=5.38(10,419+125)=56,726 lbs. 
T2=5.38(ll,784+750)=68,203 lbs. 

Perhaps attention should be called to the fact that in this 
calculation no use has been made of the factor of safety specified 
above. Arbitrarily a horizontal tension was chosen equal to one- 
fourth the breaking strength, and this gave a value for the con- 
stant c upon which is based the calculation of the load and tensions 
as just given above. The factor of safety comes out 3.29 instead 
of 3.5. It is apparent that, in using the horizontal tension recom- 
m.ended in Mark's Handbook, too small a factor of safety is pro- 
vided for. Reference will be made to this discrepancy later. 

In order that we may know where the load produces the great- 
est tension, the computation must be made for several different 
positions of the load in the same manner as above. In carrying out 
this computation one of three proceedures must be followed : 

(1). We may assume the exact position of the load measured 
horizontally and vertically from one end, and the hori- 
zontal tension of the cable as was used above. 

(2). We may take the length of the cable fixed (say, the 
length found above for the load at the middle, 2715 
feet), and assume the horizontal tension the same, one- 
fourth of the breaking strength. 



20 



Tensions in Track Cabi^es and 



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Logging Skylines 21 

(3). We may take the length of the cable fixed, and assume 
the load the same as we found above for the middle 
point of the span. 

Of these three proceedures the first is the easiest, but of no 
value, because the length of the cable found in this manner will be 
different for each position of the load. In practice the length of 
the cable is generally constant while the load is being transported. 
(Although in some logging operations the cable is let down, given 
a big sag, in picking up the load). The third method will give the 
most useful result, but is the most difficult to carry out. In fact, 
it was found impossible to make this calculation until, by the second 
proceedure, the tensions and positions of the load had been found 
for a number of points. 

The exact proceedure by the second method is as follows : 

(1). Assume a value of c, say 10,409. 

(2). Assume certain values of x^ and y-^, which seem to be 
reasonable according to one's best judgment. 

(3). By equation (58) compute s^. 

(4). By the same equation compute s,, using the same value of 
c, and values of Xo and y,, which may be found by the 
following relations, 

Xi-)-X2==horizontal span. 

y2 — yi=^verticle height of higher end above lower. 

(5). The sum of s^ and s, so found should be the length of 
cable 2715 feet. If it does not, new values of x^ and yi 
must be assumed and the same proceedure gone thru 
with again. 

It is thus seen that this method is one of "trial and error" 
which involves a vast amount of work. The computations were 
carried out for seven positions or points in the span and the results 
plotted in the curves of figure 4. Abscissas for all the curves are 
values of x^, i.e., the horizontal distance of the load from the lower 
end support. In curve L the ordinates show the load necessary 
at any point in order that the horizontal tension may be constant, 
56,000 lbs., giving a constant value of c of 10,409. T^ shows the 
total tension in the cable at the lower end fastening, and T, the 
total tension at the upper end for any position of the load. 



22 



TivNSioNS IN Track Cables and 



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FIG. 5 



Logging Skylines 23 

Curve L shows the rather surprising result that the required 
load is a minimum at the center of the span. And since the load 
is larger near the ends, the tensions in the cable are larger also for 
the load near the ends. The curves of this figure are of little prac- 
tical value except as a guide in selecting values of the coordinates 
to be used in the computations made for the curves of figure 5. 

In computing the data for the curves of this latter figure, we 
assume : 

(1). That the length of the cable is constant and equal to 
2715 feet. 

(2). That the load is constant and equal to 31,950 lbs., which 
is the load (approximately) at the middle of the span 
as first found. 

The proceedure followed is one of trial and error as described 
in method 2, but with this additional requisite : SoiH-Soo must equal a 
certain constant value, viz., 5938, which is obtained by dividing the 
load by the weight per unit length of the cable, 5.38 Ibs./^ft. This 
makes the computations for these curves twice as laborious as for 
figure 4. 

As in the former figure, the abscissas of figure 5 are values of 
Xi. Curve Tn shows how the horizontal tension varies with the 
position of the load. Curves T^ and To give the tensions in the 
cable at the lower and upper ends respectively. The form of all 
these cui"ves might have been predicted from those of figure 4. 
The maximum tension at the higher end occurs when the load is at 
the center of the span. The maximum point in the other two curves 
is slightly displaced toward the lower end. 

Although these curves give the relation between the position 
of the load and the tensions for a special case, they may be safely 
taken as a guide in predicting in general the conditions of maximum 
tensions in the cable. It is evident that if the vertical distance be- 
t\veen the end supports of the cable are less than the value taken 
above (one-fourth of the span) the tension curves will be more 
nearly symmetrical, and when the two supports are at the same 
level they will be perfectly symmetrical about an ordinate thru a 
point corresponding to the center of the span. Hence for all cases 
in which the difference in level between the ends is one- fourth the 
length of the span or less, the maximum point on the tension curves 
will occur at a position corresponding to the center of the span. 



24 



Tensions in Track Cables and 



For greater differences in height between the ends, the maximum 
point may be displaced a Httle toward the lower end. But as the 
curves are rather flat on top, no great error will be involved in 
assuming that the maximum tension in each branch of the cable 
occurs when the load is at the center of the span. This conclusion 
is of the greatest importance, for as it has been shown the calcula- 
tion of the tensions is the simplest when the load is at the center of 
the span, that is when Xi=X2 ; and when this computation is made 
we may be certain that the maximum tension is known. 

Futhermore, the conclusion stated above enables us to examine 
the most favorable conditions for the use of a track cable. As was 
previously stated, the use of a horizontal tension of one-fourth the 
breaking strength is inconsistent with a large factor of safety. 
Mark's Handbook of Mechanical Engineering recommends a fac- 
tor of safety of 4. This is absolutely impossible with a horizontal 
tension of one-fourth the breaking strength. The effect of the sag 
and the horizontal tension on the load and factor of safety is 
shown in table 1. 

Table 1. 



Sag 


Horizontal 
Tension. 


Load 


1\ 


Ts 


1 Factor 

of 
1 Safety 


l/20th 


l/4th 


31,946 1 


56,726 


68,200 


3.29 


1/1 0th 


l/4th 


41,800 1 


63,150 


70,800 


1 3.17 


l/20th 


l/5.25th 


30,602 1 


43,300 


56,000 


1 4.0 


1/lOth 


l/5.25th 


33,700 


47,900 


56,000 


1 4.0 



With a sag of one-twentieth the horizontal span and a hori- 
zontal tension of one-fourth the breaking strength, the factor of 
safety is 3.29, which is lower than is advisable to use. By using a 
greater sag, one-tenth of the span, and the same horizontal tension 
the load will be increased 30%. But this decreases the factor of 
safety to 3.17. By keeping the sag one-twentieth of the span and 
making the factor of safety 4 the load is reduced by 4.2 per cent 
and the horizontal tension to a little less than one-fifth of the break- 
ing strength. If a greater sag is used, say one-tenth of the span, 
the load is greater by 5.25 percent, and the horizontal tension is 
the same. 



Logging Skylines 25 

Hence if it is desired to use a factor of safety of 4, the sag 
should be about one-tenth of the span and the horizontal tension 
one-fifth of the breaking strength. If it is necessary to keep the 
sag small because of the topography of the ground, the horizontal 
tension should be made as small as possible and still keep the load 
at a sufficient height above the ground. 

It will be observed that in the above discussion no considera- 
tion is given to the bending stress which the cable undergoes in pas- 
sing around pulleys. Of course this is an important feature of 
cable transportation and should be given proper weight in the de- 
sign of a track cable, but it is not a part of this particular problem 
which treats only of stresses due to loading. 

IV 

Directions for Computing the Load. 

In conclusion the proceedure for finding the safe load that a 
track cable will carry will be outlined. It is assumed that the fol- 
lowing data is given : 

(a) Breaking strength of the cable. 

(b) Linear density. 

(c) Horizontal span. 

(d) Vertical distance between end fastenings. 

(1) Take the horizontal tenions to be one-fifth of the break- 
ing strength and compute the parameter, c, by 

c = - (11) 

w 

(2) Take the sag at the center of the span to be one-tenth, 
(or as near that as feasible) of the span. This gives the 
ordinate y^ of the lower end. The ordinate yo of the 
upper end is y^ plus the vertical height between the two 
ends. The absicassas, x^ and x,, are equal and each equal 
to one-half of the span. 

(3) Compute the length of each part of cable between the 
load and the lower end, and between the load and the 
upper end respectively, i.e., Sj^ and So, by 



26 Tensions in Track Cables and 



sinh </) 



Vs^ — y^'=x (58) 

<!> 

(4) The constant Soi and Soo may now be computed by 

y s 

So^ — coth(/> ■ (53) 

2 2 

(5) The load is given by 

W=w(So,+So2) (46) 

(6) The constants y^ and yoo may be found by 



yo=Vc^+s^o' (42) 

(7) The tension at each end, T^ and To, respectively, may 
be computed by 

T=w(y+yo) (51) (52) 

using y^ and yoj for T^, and y, and yoo for To. Since 
yo is taken greater than y^. To is the greater tension and 
the maximum for any position of the load. 

Altho the computations required by the outline above are some- 
what longer than those called for by the approximate formulas 
given in handbooks, they are by no means laborious, and have the 
advantage of giving exactly the load zvhicJi may be carried and the 
maximum tension involved. 



SUMMARY. 

(a). When a cable is loaded at one point, the conformation 
which the cable assumes is that of two arcs of the same common 
catenary, with the point of intersection at point of loading. 

(b). The eciuations of these two arcs have been derived. 
They involve three constants or parameters : ( 1 ) c, which is the 
parameter of the common catenary, of which the two arcs are part, 
and which depends implicitly upon the horizontal tension and linear 
density of the cable; (2) So, which is ec|ual to the length of an arc 
of the common catenary extending from the lowest point of the 
common catenary to the point of intersection (the loaded point) 
of the two arcs which represent the actual conformation of the 
cable, and which may be called the "fictitious arc" and depends 
upon the load and the linear density of the cable; (3) y,,, which is 
dependent upon the other two parameters, and is the vertical dis- 
tance of the loaded point above the directrix of the common 
catenary. 

(c). Expressions for the horizontal tension, vertical tension 
and total tension at any point in the cable have been derived. 

(d). It has been shown that in either one of the arcs the 
point of maximum tension is the highest point, i.e., the point of 
fastening at the end, and that in most cases the higher end of the 
c.?ble has the greater tension. 

(e). It has been shown for a special case that the maximum 
tension occurs in the cable when the load is at the center of the 
span, and that the point of maximum tension is the higher end of 
the cable. Furthermore, it has been shown that very probably the 
maximum tension ahvays occurs, in a catenary loaded at one point, 
when the load is at the center of the span. 

(f). A method of computing exactly the load and tensions in a 
given track cable has been outlined in section IV. To the logging 
engineer this will probably prove the most interesting and useful 
part of the paper. 



73 78 5 








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